Zark Bedalov - Practical Power Plant Engineering

In case of a failure of one of the transformers, the faulty transformer (Assume: T12 on B bus) will be isolated by being tripped automatically or manually (locally). Once the incoming breaker for the transformer T12 opens, a signal is sent to fully isolate T12 by tripping its HV breaker. This will be followed by closing the 13.8 kV bus tie breaker.

Please note, all the B buses from 13.8 kV down to 480 V are temporarily shut down following the isolation of the faulty T12 transformer. When the tie breaker closes, the healthy transformer (T11) now feeds both Buses A and B. The control system now starts restoring the power (incoming and tie breakers) sequentially from the top bus to the lowest bus in that order to operate the plant on the single transformer. The whole process of power restoration is completed within several seconds.

This automatic switching process explained above is more applicable to the power plants, which may have three or four buses operated from 13.8 kV down to 480 V. The industrial plants have fewer plant bus levels, and the switching following a transformer failure may be manual.

The actual switching is arranged on the “Break before Make” principle. This is also called a “Dead Transfer.” Therefore, the two incomers + tie breakers operate on 2 out of 3 closed principle. However, since we are also planning to connect the standby DG to the Bus B in case of a total outage, the operating logic must accommodate an operating condition of 2 out of 4 breakers closed. Should there be a total power outage, both switchgear incomer breakers are opened, a signal is sent to the standby DG to start and close onto the 13.8 kV bus. That done, the tie breaker closes next (2 out of 4 conditions) and the essential power restoration commences across the whole 13.8 kV bus and the plant.

The breaker operating and closing coils as well as the protection and trip circuits operate on 125 V dccircuits fed from the station battery. By having the DC system available, the main breakers can be operated during a total plant outage.

The plant restoration following a total blackout and DG operation follows by the appearance of voltage on the HV side of the main transformers. HV breakers are closed and a proper safe moment is awaited to initiate the restoration of the overall power distribution system.

Switching and restoring the service when the faulty transformer is brought back follows a similar procedure in reverse order. The incomer B is switched ON, on its HV side. This initiates the tie breaker T to open first, followed by closing the incomer breaker B. During this restoration to normal operation, Bus B again passes through a “dead” transition, and its feeder breakers and contactors are turned OFF during a brief transition. Latched feeder breakers are restored immediately, while the other loads must be restored by the plant control system by reviewing the permissive logic of each circuit. If the switchgear breaker cell control switches are left on Loc/Rem, they must be turned to Remote to allow the automatic reintegration of the lost services to proceed.

Automatic switching from the control room is possible only if the breaker Loc/Rem switches are held on “Remote.” All the breakers and controllers are provided with means of operating from a local and remote position. The breakers that are left on Loc position (most likely intentionally) will not be restored. The means of communication from the plant control system to the switchgear is by Ethernet (see Chapter 17).

Breaker Loc/Rem switch operation:

Loc (Local) means manual.

Rem (Remote) means automatic from control room.

Every effort shall be made to minimize the number of transformations from the grid to the loads. Unnecessary transformers add to the voltage drops as well as to the cost of the plant in the transformers and additional associated switching and protection equipment.

For this plant, we can use two or maximum three levels of transformation, as follows:

Grid → 230 kV–T1–13.8 kV → 13.2 kV–T2–4.16 kV → 4.0 kV Motors, or

Grid → 230 kV–T1–13.8 kV → 13.2 kV–T2–0.48 kV → 0.46 kV Motors and feeders

Grid → 230 kV–T1–13.8 kV → 13.2 kV–T2–4.16 kV → 4.16 kV–T3–0.48 kV → 0.46 kV Loads

Grid → 230 kV–T1–13.8 kV → 13.2 kV–T2–0.48 kV → 0.48 kV–T3–0.48 kV → 0.48 kV Ltg. Panels 1

The full range of load regulation of a power transformer is the change in secondary voltage V s, expressed in percentage of rated V s for a specified power factor. This occurs when the rated MVA output at a specified power factor is reduced to zero, with V p maintained constant. One can use these approximate formulae to calculate Δ V at any operating load and power factor:

(2.6)

For instance: Transformer: 30 MVA as a MVA b= 1 pu, having Z = 9% (0.09 pu) impedance

on 30 MVA base, efficiency 99.5% at rated load.

Total loss = (1 − 0.995) × 30 MVA = 150 kW. The total loss includes no load and load losses.

With this information, we can calculate transformer resistance R , followed by calculating the reactance X . After that, we calculate the voltage regulation.

Transformer resistance R is calculated from the load losses. Since we do not have that figure we will assume that the load loss is equal to 80% of total loss . The rest is the no load loss.

Assume operating load of 25 MVA (MVA load pu = 0.833 pu) at power factor pf = 0.85.

R or Load loss in % = 0.8 × 100 = 0.4%. Z is given as 9%

Pload = MVApu × cos ϕ = 0.833 × 0.85 = 0.71 pu

Qload = MVApu × sin ϕ = 0.833 × 0.52 = 0.43 pu

ΔVpu = (0.004 × 0.71 + 0.0899 × 0.43) = 0.0414 pu or 4.1 % . From Eq. (2.6).

The plant will include a number of large motors, for which cable sizingmust be verified with respect to the voltage drop during its start. During the motor start, the voltage at the motor terminals must be >85% of the motor nominal voltage. Here is a quick check without going through a computer study.

For selecting the power cable for the motor we use National Electrical Code (NEC) for ampacities of Cu and Al cables [1]. Most of the engineers would have this booklet (Code) on their desks and use it for the various engineering activities ranging from the switchgear and cable selection to fire protection regulations. A similar Code is also available in Canada.

Let us calculate voltage drop for a motor rated as follows:

Motor: 100 kW, 0.85 pf, 480 V

Motor sub‐transient Impedance: Zm = 17%,

Power cable length: 100 m.

Select the cable. Calculate motor nominal current:

Select the cable from NEC Ampacity table.

We look for a cable size for: 1.25 × 125 A = 155 A (∼25% margin was added).

Cable selected from the code: 3c # 1/0, Cu, 90 °C, capable of carrying 170 A.

We calculate the voltage drop on the motor kW base (kWb): 100 kW = 1 pu.

Motor impedance on the motor base if not known can be assumed as: Z m= 0.17 pu as per ANSI.

Cable impedance Z Ωfor cable 3c #1/0 AWG = 0.035 Ω/100 m. Value was taken from relevant tables.

Calculate impedance Zc pu on per unit (pu) value for 100 m cable:

Cable impedance in pu: .